Please answer correctly only

Mathematics

1 answer:

Leokris [45]3 years ago

4 0

The length of line segment EC is 35

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<h2>Event A:</h2><h3>Pairs that make the sum greater than 8:</h3>

(3,6) (4,6) (5,6) (6,6) (4,5) (5,5)

<u>Each of these pairs is </u><u>equiprobable</u><u> and which i will be using throughout the exercise</u>:

$p(pair) = \frac{1}{6} \times \frac{1}{6} = \frac{1}{36}$

$p(a) = (\frac{1}{36} \times 2) \times 4 + \frac{1}{36} \times 2 = \frac{5}{18}$

The 2 (2!) is to account for permutations like ( 3 , 6 ) and ( 6 , 3 ) and the 4 are the different combinations that are composed of different numbers.

The 2 (2!) is to account for permutations like ( 3 , 6 ) and ( 6 , 3 ) and the 4 are the different combinations that are composed of different numbers.Whereas combinations like ( 5 , 5 ) cannot be switched in order, so we do not multiply by 2!

<h2>Event B</h2><h3>Pairs that yield a sum divisible by 4 or 5: 0,4,5,8,10,12,</h3>

(1,3) (2,2) (2,6) (4,4) (5,3) (6,6) (1,4) (2,3) (5,5) (4,6)

$p(not \: b) = ( \frac{1}{36} \times 2) \times 6 + \frac{1}{36} \times 4 = \frac{4}{9}$